If the temperature of a gas increased from 100 K to 200 K and the volume of a gas decreased from 20 L to 10 L, what is the new pressure if the original pressure was kPa?

320pka

Explanation:

This is a complete question;

If the temperature of a gas increased from 100 K to 200 K and the volume of a gas decreased from 20 L to 10 L, what is the new pressure if the original pressure was 80kPa?

EXPLANATION;

Gas laws can be regarded as aws which relate the pressure, volume, as well as temperature of a gas. It can be expressed as

P1V1/T1 = P2V2/T2

Where

V1= initial volume= 20L

V2 = final volume= 10L

T1 = initial temperature= 100K

T2 = final temperature= 200K

P2=final pressure=

P1= initial pressure= 80pka

Making P2 subject of the formula from the eqn, we have

P2= (P1*V1*T2=)/(T1*V2)

If we substitute the values we have

P2= (80 × 20×200)/(100×10)

P2= 320pka

Hence, the new pressure is 320pka

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